Question: $ g(x) = \int_{\,0}^{\,x}\sqrt{5+4\cos {t}}\,dt \,$ $ g\,^\prime(\pi)\, = $
Answer: The Fundamental Theorem of Calculus If $~ g(x)=\int_a^xf(t)\,dt\,$, then $~g^\prime (x)=f(x)\,$ This only works if $f$ is continuous on $[a,b]$. Thankfully, the function $f(t) = \sqrt{5+4\cos {t}}$ is continuous on $[0,\pi]$. Applying the theorem We're given: $ g(x) = \int_{\,0}^{\,x}\sqrt{5+4\cos{t}}\,dt $ So the theorem tells us: $ g\,^\prime(x) =\sqrt{5+4\cos{x}}$ Evaluating $g'(\pi)$ $ g\,^\prime(\pi) = \sqrt{5+4\cdot \cos{\pi}}=\sqrt{5+4\cdot (-1)}=\sqrt{5-4}=1$ The answer: $g'(\pi)=1$